Difference between revisions of "2013 AMC 10A Problems/Problem 13"

(Video Solution)
 
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Summing, we get <math>40 + 8 + 6 + 4 + 2 = \boxed{\textbf{(B) }60}</math>
 
Summing, we get <math>40 + 8 + 6 + 4 + 2 = \boxed{\textbf{(B) }60}</math>
 
==Video Solution==
 
https://youtu.be/YXBC8gCVnNE
 
 
~savannahsolver
 
  
 
==See Also==
 
==See Also==

Latest revision as of 10:15, 21 January 2021

Problem

How many three-digit numbers are not divisible by $5$, have digits that sum to less than $20$, and have the first digit equal to the third digit?


$\textbf{(A)}\ 52 \qquad\textbf{(B)}\ 60  \qquad\textbf{(C)}\ 66 \qquad\textbf{(D)}\ 68 \qquad\textbf{(E)}\ 70$

Solution

We use a casework approach to solve the problem. These three digit numbers are of the form $\overline{xyx}$.($\overline{abc}$ denotes the number $100a+10b+c$). We see that $x\neq 0$ and $x\neq 5$, as $x=0$ does not yield a three-digit integer and $x=5$ yields a number divisible by 5.

The second condition is that the sum $2x+y<20$. When $x$ is $1$, $2$, $3$, or $4$, $y$ can be any digit from $0$ to $9$, as $2x<10$. This yields $10(4) = 40$ numbers.

When $x=6$, we see that $12+y<20$ so $y<8$. This yields $8$ more numbers.

When $x=7$, $14+y<20$ so $y<6$. This yields $6$ more numbers.

When $x=8$, $16+y<20$ so $y<4$. This yields $4$ more numbers.

When $x=9$, $18+y<20$ so $y<2$. This yields $2$ more numbers.

Summing, we get $40 + 8 + 6 + 4 + 2 = \boxed{\textbf{(B) }60}$

See Also

2013 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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